Question

A sample of 114 mortgages approved during the current year showed that 36 were issued to a single-earner family or individual. The historical percentage is 28 percent. At the .05 level of significance in a right-tailed test, has the percentage of single-earner or individual mortgages risen?
(a-1) H0: π ≤ .28 versus H1: π > .28. Choose the right option. Reject H0 if zcalc > 1.645 Reject H0 if zcalc < 1.645 a b
(a-2) Calculate the test statistic. (Round intermediate calculations to 2 decimal places. Round your answer to 4 decimal places.) Test statistic
(a-3) The null hypothesis should be rejected.
i. True
ii. False

(b) Is this a close decision?
i. Yes
ii. No

(c) State any assumptions that are required.
i. n π < 10 and n(1 − π ) > 10
ii. n π < 10 and n(1 − π ) < 10
iii. n π > 10 and n(1 − π ) > 10
iv. n π > 10 and n(1 − π ) < 10

Answer 1

Answer:

a-1) Reject H0 if zcalc > 1.645

a-2) $z=\frac{0.316 -0.28}{\sqrt{\frac{0.28(1-0.28)}{114}}}=0.8561$  

a-3) ii. False

b) ii. No

c) iii. n π > 10 and n(1 − π ) > 10

Step-by-step explanation:

1) Data given and notation

n=114 represent the random sample taken

X=36 represent the people that were issued to a single-earner family or individual

$\hat p=\frac{36}{114}=0.316$ estimated proportion of people that were issued to a single-earner family or individual

$p_o=0.28$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)  

(a-1) H0: π ≤ .28 versus H1: π > .28. Choose the right option. Reject H0 if zcalc > 1.645 Reject H0 if zcalc < 1.645 a b

We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.28.:  

Null hypothesis:$p\geq 0.28$  

Alternative hypothesis:$p < 0.28$  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

The rejection zone would be on this case :

Reject H0 if zcalc > 1.645

Since is a right tailed test

(a-2) Calculate the test statistic. (Round intermediate calculations to 2 decimal places. Round your answer to 4 decimal places.) Test statistic

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.316 -0.28}{\sqrt{\frac{0.28(1-0.28)}{114}}}=0.8561$  

(a-3) The null hypothesis should be rejected.

False, since our calculated value is less than our critical value we Fails to reject the null hypothesis

(b) Is this a close decision?

False the calculated value is significantly less than the critical value so we FAIL to reject the null hypothesis with enough confidence.

(c) State any assumptions that are required.

In order to satisfy the conditions we need the following two requirements:

iii. n π > 10 and n(1 − π ) > 10

And are satisfied:

114*0.28=31.92>10

114(1-0.28)=82,08>10